For 36 3 6 13. For 48 4 8 24. I think there are more but I can't think of them
Answer:
This question requires a comparison between two different variables 6% and 16% values. If we let x=$ loaned on 6% loans and y=$ loaned on 16% loans, then we can relate two equations.
0.06x + 0.16y = $1500 --> referencing the interest earned from each percentage loaned.
x + y = $16000 --> referencing the total amount of money loaned out.
Rearrange either equation and substitute for a value in the other equation or use elimination to determine each individual variable.
Step-by-step explanation:
To find the z-score for a weight of 196 oz., use

A table for the cumulative distribution function for the normal distribution (see picture) gives the area 0.9772 BELOW the z-score z = 2. Carl is wondering about the percentage of boxes with weights ABOVE z = 2. The total area under the normal curve is 1, so subtract .9772 from 1.0000.
1.0000 - .9772 = 0.0228, so about 2.3% of the boxes will weigh more than 196 oz.
Answer:
0.2805
Step-by-step explanation:
Given that p = 20% = 0.2, n = 300.
The mean (μ) = np = 0.2 * 300 = 60
The standard deviation (σ) = 
The z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:

For x = 57:

For x = 62:

From the normal distribution table, P(57 < x < 62) = P(-0.43 < z < 0.29) = P(z < 0.29) - P(z < -0.43) = 0.6141 - 0.3336 = 0.2805
Not really sure whats up with that inequality but she can't, if she has to spend 1.5 hours in a lab that leaves her with 5 hours. 5/4 is 1.25, so no, she can only spend 1.25 hours with each student.